Solve 4x^3-27x^2+24x+28=0

$$4 x ^ { 3 } - 27 x ^ { 2 } + 24 x + 28 = 0$$

Solve for x

$$x = \frac{3 \sqrt{65} + 19}{8} \approx 5.398346656$$

$$x=\frac{19-3\sqrt{65}}{8}\approx -0.648346656$$

$$x=2$$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 28 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±7,±14,±28,±\frac{7}{2},±\frac{7}{4},±1,±2,±4,±\frac{1}{2},±\frac{1}{4}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide 4x^{3}-27x^{2}+24x+28 by x-2 to get 4x^{2}-19x-14. Solve the equation where the result equals to 0.

$$4x^{2}-19x-14=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -19 for b, and -14 for c in the quadratic formula.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 4\left(-14\right)}}{2\times 4}$$

Do the calculations.

$$x=\frac{19±3\sqrt{65}}{8}$$

Solve the equation 4x^{2}-19x-14=0 when ± is plus and when ± is minus.

$$x=\frac{19-3\sqrt{65}}{8} x=\frac{3\sqrt{65}+19}{8}$$

List all found solutions.

$$x=2 x=\frac{19-3\sqrt{65}}{8} x=\frac{3\sqrt{65}+19}{8}$$